Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 3 HCF and LCM Ex 3.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 5 Maths Chapter 3 HCF and LCM Ex 3.3

Question 1.

Find LCM of the following :

(a) 5, 10

(b) 6,18

(c) 25, 50

(d) 9, 24

Solution:

(a) Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, …….

Multiples of 10 = 10, 20, 30, 40, 50, ……..

Common multiples of 5 and 10 = 10, 20, 30, 40, 50, ……….

The lowest common multiple = 10

So, LCM of 5 and 10 = 10

(b) Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, …….

Multiples of 18 = 18, 36, 54, ……….

Common multiples of 6 and 18 = 18, 36, 54, ………

The lowest common multiple = 18

So, LCM of 6 and 18= 18

(c) Multiples of 25 = 25, 50, 75, 100, 125, 150, …….., …….

Multiples of 50 = 50, 100, 150, 200, ……, ……, ……..

Common multiples of 25 and 50= 50, 100, 150, ………..

The lowest common multiple = 50

So, LCM of 25 and 50 = 50

(d) Multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, ………

Multiples of 24 = 24, 48, 72, 96, …., ….., …….

Common multiples of 9 and 24 = 72,

So, LCM of 9 and 24 = 72

Question 2.

Find LCM of the follówing:

(a) 4, 8 and 12

(b) 6, 12 and 24

(e) 15, 18 and 27

(d) 24, 36 and 40

Solution:

(a) Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36,. 40, 44, 48,

Multiples of 8 = 8, 16, 24, 32, 40, 48, …….

Multiples of 12 = 12, 24, 36, 48, 60, ……..

Common multiples of 4, 8 and 12 = 24, 48

The lowest common multiple = 24

So, LCM of 4, 8, 12 = 24

(b) Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, ……

Multiples of 12 = 12, 24, 36, 48, 60, 72, …….

Multiples of 24 = 24, 48, 72, 96, ……..

Common multiples of 6, 12 and 24 = 24, 48, 72, …..

The lowest common multiple = 24

So, LCM of 6, 12, 24 = 24

LCM of 15, 18 and 27 = 3 × 3 × 5 × 2 × 3 = 270

LCM of 24, 36 and 40 = 2 × 2 × 2 × 3 × 3 × 5 = 360

Question 3.

Find LCM of following using Prime factorisation :

(a) 32,40

(b) 24, 36

(c) 15, 30 and 45

(d) 40, 44 and 48

Solution:

(a)

32 = 2 × 2 × 2 × 2 × 2

40 = 2 × 2 × 2 × 5

Common factors = 2 × 2 × 2

Remaining factors = 2 × 2 × 5

So, LCM = 2 × 2 × 2 × 2 × 2 × 5 = 160

(b) 24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3×3

Common factors = 2 × 2 × 3

Remaining factors = 2 × 3

So, LCM = 2 × 2 × 2 × 3 × 3 = 72

(c) 15 = 3 × 5

30 = 2 × 3 × 5

45 = 3 × 3 × 5

Common factors = 3 × 5

Remaining factors = 2 × 3

So, LCM = 3 × 5 × 2 × 3 = 90

(d) 40 = 2 × 2 × 2 × 5

44 = 2 × 2 × 11

48 = 2 × 2 × 2 × 2 × 3

Common factors = 2 × 2

Remaining factors = 5 × 11 × 2 × 2 × 3

So, LCM = 2 × 2 × 2 × 2 × 5 × 3 × 11

= 2640

Question 4.

Find LCM of following using Division method :

(a) 15, 20

(b) 12, 38

(c) 30, 45 and 50

(d) 40, 68 and 60

Solution:

(a)

LCM of 15 and 20 = 2 × 2 × 5 × 3 = 60

(b)

LCM of 12 and 38 = 2 × 2 × 3 × 19 = 228

(c)

LCM of 30, 45 and 50 = 2 × 3 × 3 × 5 × 5 = 450

(d)

LCM of 40, 68 and 60 = 2 × 2 × 2 × 3 × 5 × 17 = 2040

Question 5.

Find the smallest number which is divisible by 12, 15 and 20 completely ?

Solution:

LCM of 12, 15 and 20

LCM = 2 × 2 × 3 × 5 = 60

Smallest number which is divisible by 12, 15 and 20 = 60

Question 6.

One child jumps 3 feet high and another jumps 4 feet high. If both the children continue jumping together in same direction then after how many feet they will be together again ?

Solution:

We are to find the LCM of 3 and 4

LCM = 3 × 4 = 12

They will be together again after 12 feet.

Question 7.

How many minimum number of students are required from a class to make groups of 4 each and 5 each so that no student is left ?

Solution:

We are to find the LCM of 4 and 5

LCM = 4 × 5 = 20

Number of students required = 20

Question 8.

Three bells ring with a time gap of 10 min, 20 min and 30 min respectively in a school. If all bells are rung together at 8:00 am then after how long the beUs would ring together again ?

Solution:

We are to find the LCM of 10, 20 and 30

LCM = 2 × 5 × 2 × 3 = 60

The bells will ring together after 60 min. i.e. 1 hours.

Thus, the bells will ring together at 9:00 am